Prove that there are exactly 16 solutions to this problem.

Question

Show that are are only 16 integer solutions to the following equation: $$11x + 8y + 17 = xy$$ What I tried: I took a modulo 2, and I got that $y$ must be even and $x$ must be odd. But beyond that, I don't really know how to start.

Answer 1

Here is an algebraic method: $$17=xy-11x-8y$$ $$\iff 105=xy-11x-8y+88$$ $$\iff 105=(x-8)(y-11)$$

Answer 2

Hint:

1. Solve for $x$ to see, that there is an integer solution for a given $y\in\mathbb Z$, iff: $$y-11\mid 8y+17$$
2. Conclude $y-11\mid 105$.
3. Check the remaining (finitely many) possibilites.

Answer 3

11x + 17 = y(x - 8)
y = (11x + 17)/(x - 8)
y = 11(x - 8) + {105/(x - 8)}

Now, 11(x - 8) can have infinite x. But, 105/(x - 8) should be an integer too for y to be an integer. So, we see that 105 is divisible by these numbers = 1 x 105, 3 x 35, 5 x 21, 7 x 15. So, 105 is divisible by 1,3,5,7,15,21,35,105. Also, negative of these numbers. So we have 8 + 8 = 16 numbers.

So, possible x will be solved for like this:

x - 8 = 1
x - 8 = 3
...
x - 8 = 105
x - 8 = -1
x - 8 = -3
...
x - 8 = -105

These are the values of x for which y will be an integer! (16)